优化linux内核pick_next_task函数

来源：互联网发布：最优化求非线性方程编辑：程序博客网时间：2024/05/01 16:09

linux3.3.7的pick_next_task函数的源码如下：

static inline struct task_struct *pick_next_task(struct rq *rq){const struct sched_class *class;struct task_struct *p;/* * Optimization: we know that if all tasks are in * the fair class we can call that function directly: */if (likely(rq->nr_running == rq->cfs.h_nr_running)) {p = fair_sched_class.pick_next_task(rq);if (likely(p))return p;}for_each_class(class) {p = class->pick_next_task(rq);if (p)return p;}BUG(); /* the idle class will always have a runnable task */}

仔细分析该代码，发现在rq->nr_running==0的时候，并不是最优的代码，fair_sched_class.pick_next_task会被执行两次，stop_sched_class.pick_next_task以及rt_sched_class.pick_next_task会各被执行一次，最后才能执行到idle_sched_class.pick_next_task。那么如何进行优化呢？

static inline struct task_struct *pick_next_task(struct rq *rq){        struct task_struct *p;        if(rq->nr_running!=0){                p = stop_sched_class.pick_next_task(rq);                if(p)                        return p;                if(rq->rt.rt_nr_running){                        p = rt_sched_class.pick_next_task(rq);                        if(p)                                return p;                }                if(rq->cfs.nr_running){                        p = fair_sched_class.pick_next_task(rq);                        if(p)                                return p;                }        }        else{                p = idle_sched_class.pick_next_task(rq);                if(p)                        return p;        }        BUG(); /* the idle class will always have a runnable task */}

改为上面的代码后，当rq->nr_running==0的时候便可以直接选择idle_sched_class.pick_next_task，但是不知道有没有什么副作用？